Multiplicative order (${\mathrm{ord}}_n(a)$)

For coprime integers $a,n$ with $n>1$, the multiplicative order is the least positive integer $d$ such thatIt counts the distinct powers in the cyclic subgroup generated by the congruence class of $a$. Thus it is an order of a group element in the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$.^[ref]

For example, the modular exponentiation $2^k\mathrm{mod}7$, starting at $k=0$, produces $1,2,4,1,2,4,\dots$. The first return to one takes three steps, so ${\mathrm{ord}}_7(2)=3$. Starting at the congruence class three and repeatedly multiplying by two instead gives $3,6,5,3,\dots$: these are the other three units modulo seven, outside the cyclic subgroup generated by two.

Euler's theorem guarantees a return to one at exponent $\phi (n)$, where $\phi$ is Euler's totient function. Thus the multiplicative order exists for every unit.^[ref] In fact,By Lagrange's theorem, the size of the cyclic subgroup divides the size of the unit group, which is Euler's totient function. For two modulo seven, this is $3\mid 6$; the multiplicative order need not equal Euler's totient function. If $a\equiv 1(\mathrm{mod}n)$, its multiplicative order is one.

The coprime integers hypothesis is necessary: if $a^d=1+kn$, any common divisor of $a$ and $n$ divides one. For instance, powers of two modulo eight give $1,2,4,0,0,\dots$, with no positive-exponent return to one. This sequence eventually settles, but two has no multiplicative order modulo eight.

For a reduced fraction $p/q$ with $0<p<q$, $q>1$, and $q,10$ coprime integers, its decimal representation repeats immediately after the decimal point. The shortest repeating block has lengthHere $\mathrm{ord}$ denotes multiplicative order, and $\phi$ denotes Euler's totient function. Changing the numerator preserves the decimal period as long as the fraction remains reduced.^[ref]

For example, long division of $1/7$ gives the remainders $1,3,2,6,4,5,1$, with emitted digits $1,4,2,8,5,7$. ConsequentlyEach remainder determines both the next digit and the next remainder; returning to one restarts the same six steps.

More generally, initialize long division with $r_0=p$. At each step, division with remainder givesso $r_j\equiv p10^j(\mathrm{mod}q)$. Since $p,q$ are coprime integers, the modular inverse of $p$ shows that $r_L=r_0$ exactly when $10^L\equiv 1(\mathrm{mod}q)$. To see why the digit block cannot repeat sooner than the remainders, a purely repeating decimal with a block of $L$ digits represents $D/(10^L-1)$ for an integer $D$. A reduced fraction $p/q$ with that value forces $q\mid 10^L-1$. Hence the shortest digit repeat and the multiplicative order coincide.

Knowing the decimal period predicts how many digits must be computed and stored: one block determines the whole repeating decimal. It also constructs the block directly. With $L={\mathrm{ord}}_q(10)$, the integeris the block, padded on the left with zeros to $L$ digits. For $1/11$, $L=2$ and $D=9$, so the block is $09$ and $1/11=0.\overline{09}$. Reduction matters: $7/21=1/3=0.\overline{3}$ has decimal period one, whereas ${\mathrm{ord}}_{21}(10)=6$.

If the reduced denominator contains prime factors two or five, ten has no multiplicative order modulo that whole denominator. WriteThe decimal preperiod has exactly $k_0$ digits. If $m>1$, the remaining repeating decimal has decimal period ${\mathrm{ord}}_m(10)$. For instance, $28=2^2\cdot 7$ gives $1/28=0.03\overline{571428}$: two prefix digits, then six repeating digits. If $m=1$, the decimal representation is a terminating decimal with exactly $k_0$ fractional places. We use the representation ending in zeros, rather than its alternative ending in nines; for example $1/40=0.025$ terminates after three places.

A short derivation explains both cases. After $k$ digits, multiplying the reduced fraction by $10^k$ and discarding its integer part leaves a reduced fraction whose denominator is $q/\gcd (q,10^k)$. This denominator first becomes coprime to ten at $k=k_0$. It then equals $m$, giving an immediate repeat when $m>1$, or an integer when $m=1$. The same long division argument works for positional notation in any base $b\ge 2$: for a reduced denominator $q>1$ coprime to $b$, the shortest repeating block has length ${\mathrm{ord}}_q(b)$.

For an $L$-digit decimal integer $n$, moving its leading digit to the end gives a cyclic rotation $c_1(n)$ satisfyingThus, when a modulus $d$ divides $10^L-1$, every rotation has the same divisibility by $d$ as $n$, up to multiplication by a power of ten.

For a unit $a$ modulo $N$, the function $x\mapsto a^x\mathrm{mod}N$ is periodic on the integers.

Let $a,n$ be coprime integers with $n>1$, and let $d={\mathrm{ord}}_n(a)$ be their multiplicative order. For integers $x,y$,Negative exponents mean powers of the modular inverse of $a$. Thus periodic modular exponentiation has smallest positive period $d$, and $1,a,\dots ,a^{d-1}$ are distinct congruence classes.^[ref]

This lets a huge modular exponentiation reuse a short list: reduce its exponent modulo the multiplicative order, then look up the corresponding power. For example, $1000\equiv 1(\mathrm{mod}3)$ gives $2^{1000}\equiv 2(\mathrm{mod}7)$, while $1001\equiv 2(\mathrm{mod}3)$ gives $2^{1001}\equiv 4(\mathrm{mod}7)$. When the multiplicative order is unknown, Euler's theorem permits reduction modulo Euler's totient function, though that period may be longer. Repeated squaring remains useful when computing either the unreduced or the reduced power.

For a proof outline, multiply $a^x\equiv a^y$ by the modular inverse of $a^y$, leaving $a^{x-y}\equiv 1$. Apply Euclidean division to write $x-y=td+r$ with $0\le r<d$. Then $a^{x-y}\equiv a^r$; minimality of the multiplicative order forces $r=0$. The reverse implication follows from $a^d\equiv 1$. The coprime integers hypothesis enables cancellation: modulo eight, $2^3\equiv 2^4\equiv 0$, but this repetition supplies no multiplicative order for two.

To verify a candidate order $r$, check $a^r\equiv 1$ and rule out $a^{r/q}\equiv 1$ for every prime divisor $q$ of $r$.